Take an example of the below plot for an LPF (Source : WikiPedia)

enter image description here

The plot starts from $0$. We know that the fourier transform of any signal brings in negative frequencies due to complex exponentials (which are just as real as positive frequencies, just to be clear) but most of the frequency response plots I have observed for practical causal filters, they start from $0$.

Indeed, the frequency response of an ideal LPF is a rectangular function symmetric about the $jw$ axis and the impulse response is a non-causal sinc function.

I have the following questions: Why do we represent/plot the frequency response of a causal filter with only positive frequency plane? Does it have to do something with making the non-causal ideal sinc filter, a causal one? (But don't we have to make the impulse response [time domain] causal in order to achieve that? [although I don't know what's the frequency response of a causal sinc]) or is it understood that there is a symmetric negative frequency part and we just ignore it?


1 Answer 1


This has absolutely nothing to do with causality. The frequency response of a real-valued filter (i.e., one with a real-valued impulse response) is (conjugate) symmetric, i.e., the negative frequencies are redundant. That's why it is sufficient to show the frequency response at non-negative frequencies only.

You can easily see that symmetry as follows. The frequency response of a system with impulse response $h(t)$ is given by

$$H(\omega)=\int_{-\infty}^{\infty}h(t)e^{-j\omega t}dt\tag{1}$$

If $h(t)$ is real-valued, taking the complex conjugate of $(1)$ gives

$$H^*(\omega)=\int_{-\infty}^{\infty}h^*(t)e^{j\omega t}dt=\int_{-\infty}^{\infty}h(t)e^{j\omega t}dt=H(-\omega)\tag{2}$$

So, for real-valued $h(t)$, we have


With $H(\omega)=|H(\omega)|e^{j\phi(\omega)}$ we get


i.e., the magnitude of the frequency response is an even function of frequency, and


i.e., the phase is an odd function of frequency.

The same is true for discrete-time systems.

  • $\begingroup$ Yes, this was amongst one of conclusions I had but I wasn't sure. Thanks a lot for the confirmation! It's always a pleasure to read your answers :-) $\endgroup$
    – paulplusx
    Nov 9, 2018 at 13:12

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